Mathematical Reasoning - Online Test

Q1. The logically equivalent propositions of p↔q is
Answer : Option C
Explaination / Solution:

p↔q≡(p→q)∧(q→p)

Q2. ∼(p∨q)∨(∼p∧q) is logically equivalent to
Answer : Option A
Explaination / Solution:


 distributive law

    since 

  since 


Q3. If the inverse of implication p→q is defined as ∼p→∼q , then the inverse of proposition (p∧∼q)→r is
Answer : Option C
Explaination / Solution:

as given the rule of inverse in question it becomes ∼(p∧∼q)→∼r =(∼p∨q)→∼r

Q4. Which of the following is logically equivalent to (p∧q) ?
Answer : Option B
Explaination / Solution:

∼(∼(p∧q)) since∼p∨∼q≡∼(p∧q) ∼(∼p)≡p

Q5. Let p and q be two propositions. Then, the contrapositive of the implication p→q is
Answer : Option B
Explaination / Solution:

the contrapositve of p→q is∼q→∼p

Q6. Negation of the statement ∼p→(q∨r) is
Answer : Option A
Explaination / Solution:

rules of negation ∼(p→q)≡p∧∼q Hence ∼p∧(∼q∧∼r)

Q7. ∼p∨∼q is logically equivalent to
Answer : Option B
Explaination / Solution:

The answer is rule of negation for ∼(p→∼q)≡∼p∨∼q

Q8. ∼(p∧q) is logically equivalent to
Answer : Option B
Explaination / Solution:

∼(p∧q)≡∼p∨∼q De Morgan's law

Q9. Which of the following is logically equivalent to p↔q ?
Answer : Option B
Explaination / Solution:

By definition p↔q=(p→q)∧(q→p)

Q10. If p→(q∨r) is false , then the truth values of p , q , r are respectively
Answer : Option B
Explaination / Solution:

 = F



qVr=F implies both q and r must be F.HenceT,F,F